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Let $1<Y<2$. Why is the following true? \begin{align*} \lim_{\delta\to 0}\int_{\delta}^1x^{-Y}e^{-x/\delta}dx=0 \end{align*}

Johannes
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  • i feel it's obvious (for me ) :-) . – what'sup Oct 07 '13 at 22:51
  • Do you know under what conditions you are able to interchange the limit with integral? – Mhenni Benghorbal Oct 07 '13 at 23:04
  • According to Mathematica, when $Y = 3/2$ your integral is equal to $$-2e^{-1/\delta}+\frac{\frac{2}{e}+2\sqrt{\pi } \operatorname{erf}(1)-2\sqrt{\pi} \operatorname{erf}\left(\frac{1}{\sqrt{\delta}}\right)}{\sqrt{\delta}}$$ which tends to $+\infty$ as $\delta \to 0^+$, not zero. Here $\operatorname{erf}$ is the error function. Maybe there is an error in the question? – Antonio Vargas Oct 07 '13 at 23:39
  • @what'sup, perhaps you should take a second look. – Antonio Vargas Oct 07 '13 at 23:40
  • Should the lower limit of integration perhaps be $\delta^{-1}$? – David H Oct 07 '13 at 23:56

1 Answers1

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I'll assume $\delta > 0$. Making the change of variables $x=\delta t$ yields

$$ \int_{\delta}^1 x^{-Y}e^{-x/\delta}\,dx = \delta^{1-Y} \int_1^{1/\delta} t^{-Y} e^{-t}\,dt. $$

The new integral converges as $\delta \to 0$, so we conclude that

$$ \int_{\delta}^1 x^{-Y}e^{-x/\delta}\,dx \sim \delta^{1-Y} \int_1^\infty t^{-Y} e^{-t}\,dt $$

as $\delta \to 0$. In particular, if $Y>1$ then

$$ \lim_{\delta \to 0^+} \int_{\delta}^1 x^{-Y}e^{-x/\delta}\,dx = +\infty. $$

Antonio Vargas
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  • This looks good to me. I had numerically evaluated the integral using Matlab and thought it converged, must have done something wrong. – Johannes Oct 08 '13 at 00:38